How to calculate capacitors in parallel
Capacitors are in parallel when each one connects across the same two points, so every capacitor has the same voltage across it. Unlike resistors, parallel capacitors simply add up: it is like making one capacitor with a bigger plate area. This calculator works out the total capacitance of any number of parallel capacitors and, for the voltage you enter, the charge and energy stored in each one, and shows the three-digit code printed on each part.
How to use the parallel capacitor calculator
- Enter the source voltage. It is the same across every capacitor. Leave it at 0 V if you only need the total capacitance.
- Choose how many capacitors with the + and − buttons or by typing the number, up to 500.
- Type each capacitance the way it is written on parts: 100n, 4.7u,
10uF, 22p or 0.1 (farads). The panel shows the three-digit code for each value. - Read the results under the drawing: total capacitance, total charge, stored energy and the largest share. Click a capacitor in the drawing for its own calculation.
- Open the Calculation tab for the maths written out, or the Results table for every capacitor's charge, share and energy.
The parallel capacitance formula
Each capacitor holds Q = C × V. With the same V on all of them the charges add, so the capacitances add:
Ceq = C1 + C2 + C3 + C4
Your 4 capacitors Ceq = 100 nF + 220 nF + 470 nF + 1 µF = 1.79 µF.
The total is always larger than the largest single capacitor. Equal capacitors give C × n: two 100 nF make 200 nF.
Charge and energy in each capacitor
Qk = Ck × V Ek = ½ × Ck × V²
The largest capacitor holds the largest share of the charge, Ck / Ceq. The energy is what the bank can deliver: a 1000 µF capacitor at 12 V stores 72 mJ, enough to keep a small circuit alive for a moment when the supply dips, which is why parallel banks are used for smoothing and decoupling.
Your bank, step by step
- Add the capacitances: Ceq = 100 nF + 220 nF + 470 nF + 1 µF = 1.79 µF.
- Total charge: Q = Ceq × V = 1.79 µF × 12 V = 21.48 µC; energy ½CV² = 128.9 µJ.
- Each capacitor holds Q = C × V:
| Capacitor | Charge Q = C × V | Energy ½CV² | Share |
|---|---|---|---|
| C1 = 100 nF | 1.2 µC | 7.2 µJ | 5.6% |
| C2 = 220 nF | 2.64 µC | 15.8 µJ | 12% |
| C3 = 470 nF | 5.64 µC | 33.8 µJ | 26% |
| C4 = 1 µF | 12 µC | 72 µJ | 56% |
| Total: 1.79 µF | 21.5 µC | 129 µJ | 100% |
Worked example: 100 nF, 220 nF and 1 µF at 12 V
Add them: 100 nF + 220 nF + 1 µF = 1.32 µF. The bank holds Q = 1.32 µF × 12 V = 15.84 µC and stores ½ × 1.32 µF × 12² = 95.04 µJ. The 1 µF part holds 76% of the charge; the 100 nF only 7.6%.
The three-digit code on capacitors
Small capacitors are marked with a three-digit code in picofarads: the first two digits, then a power of ten. 104 is 10 × 10⁴ pF = 100 nF; 223 is 22 × 10³ pF = 22 nF; 475 is 4.7 µF. Values under 100 pF are printed as they are (22 = 22 pF). The drawing prints each capacitor's code on its body, and the panel shows it next to the value, so you can match the part in your hand to the number in the calculator. A letter after the code is the tolerance (J = 5%, K = 10%, M = 20%).
Questions
Why do capacitors add in parallel when resistors don't?
Capacitance is plate area over plate spacing. Side by side, the areas add. Resistance is the opposite kind of quantity: more paths in parallel means less resistance, which is why resistors add their conductances instead.
Why put a small capacitor next to a big one?
Decoupling: a big electrolytic holds charge for slow dips, a small ceramic responds to fast spikes. Their capacitances add, but the point is their different speeds.
What voltage rating do I need?
Every capacitor in the bank sees the full source voltage, so each one needs a rating above it, usually 1.5 to 2 times.
What about capacitors in series?
Then the same charge sits on each and the reciprocals add, giving less than the smallest; the series capacitor calculator works those out.